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Two long straight wires P and Q carrying equal current 10 A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F_{1}   - If distance between wires is halved and currents on them are doubled, force F_{2} on 10 cm length of wire P will be: 

Option: 1

\frac{F_1}{8}


Option: 2

8 \mathrm{~F}_1


Option: 3

10 \mathrm{~F}_1


Option: 4

\frac{F_1}{10}


Answers (1)

best_answer

\begin{aligned} & \mathrm{F}=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{r}} \Rightarrow \mathrm{F}=\frac{\mu_0 \mathrm{I}^2 \ell}{2 \pi \mathrm{r}} \\ & \ell=10 \mathrm{~cm}(\text { Both }) \Rightarrow \mathrm{F} \propto \frac{\mathrm{I}^2}{\mathrm{r}} \\ & \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\left(\frac{\mathrm{I}}{2 \mathrm{I}}\right)^2\left(\frac{5 / 2}{5}\right) \Rightarrow \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{1}{8} \Rightarrow \mathrm{F}_2=8 \mathrm{~F}_1 \end{aligned}

 

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Divya Prakash Singh

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